「每日LeetCode」2020年12月9日

本文最后更新于：2023年3月19日 晚上

Lt999. 可以被一步捕获的棋子数

999. 可以被一步捕获的棋子数

在一个 8 x 8 的棋盘上，有一个白色的车（Rook），用字符 'R' 表示。棋盘上还可能存在空方块，白色的象（Bishop）以及黑色的卒（pawn），分别用字符 '.'，'B' 和 'p' 表示。不难看出，大写字符表示的是白棋，小写字符表示的是黑棋。
车按国际象棋中的规则移动。东，西，南，北四个基本方向任选其一，然后一直向选定的方向移动，直到满足下列四个条件之一：

  • 棋手选择主动停下来。
  • 棋子因到达棋盘的边缘而停下。
  • 棋子移动到某一方格来捕获位于该方格上敌方（黑色）的卒，停在该方格内。
  • 车不能进入/越过已经放有其他友方棋子（白色的象）的方格，停在友方棋子前。

你现在可以控制车移动一次，请你统计有多少敌方的卒处于你的捕获范围内（即，可以被一步捕获的棋子数）。

示例 1：

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输入：[[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".","R",".",".",".","p"],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."]]
输出：3
解释：
在本例中，车能够捕获所有的卒。

示例 2：

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输入：[[".",".",".",".",".",".",".","."],[".","p","p","p","p","p",".","."],[".","p","p","B","p","p",".","."],[".","p","B","R","B","p",".","."],[".","p","p","B","p","p",".","."],[".","p","p","p","p","p",".","."],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."]]
输出：0
解释：
象阻止了车捕获任何卒。

示例 3：

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输入：[[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".","p",".",".",".","."],["p","p",".","R",".","p","B","."],[".",".",".",".",".",".",".","."],[".",".",".","B",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".",".",".",".",".","."]]
输出：3
解释：
车可以捕获位置 b5，d6 和 f5 的卒。

提示：

  1. board.length == board[i].length == 8
  2. board[i][j] 可以是 'R'，'.'，'B' 或 'p'
  3. 只有一个格子上存在 board[i][j] == 'R'

思路

找到 R，然后上下左右直走，如果遇到 p 就 count 加一，如果遇到 R 或到边界则跳出，返回结果

解答

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/**
 * @param {character[][]} board
 * @return {number}
 */
var numRookCaptures = function (board) {
  let count = 0;
  let i = 0,
    j = 0;
  for (i = 0; i < 8; i++) {
    for (j = 0; j < 8; j++) {
      if (board[i][j] === "R") break;
    }
    if (board[i][j] === "R") break;
  }
  if (i == 8 || j == 8) return 0;
  let x = i,
    y = j;
  while (x > 0 && board[--x][y] !== "B")
    if (board[x][y] === "p") {
      count++;
      break;
    }
  x = i;
  while (x < 7 && board[++x][y] !== "B")
    if (board[x][y] === "p") {
      count++;
      break;
    }
  x = i;
  while (y > 0 && board[x][--y] !== "B")
    if (board[x][y] === "p") {
      count++;
      break;
    }
  y = j;
  while (y < 7 && board[x][++y] !== "B")
    if (board[x][y] === "p") {
      count++;
      break;
    }
  return count;
};